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# Laplace transform, pretty quick (renewal process)

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1. Laplace transform, pretty quick (renewal process)
I have a renewal function given as for .

The interarrival times are identically distributed with distribution density function which is given by the equation .

I need to show that

Having solved the second order differential equation, I get that . Clearly I just need my constants A and B, but the only possible way I can do it straight from here is by using the fact that f is a density, so . I'm not sure what to do from here.

I know that , where .

So I found

Now rearranging the aforementioned equation, I'll get ... but substitution is really tricky. I expect that I've gone wrong somewhere in my logic but I can't see why.

Edit: I guess I could just use the f(t) I'm trying to prove, find its transform and show that it gives us ...? But this is really dodgy!

Anyone have any suggestions?
Last edited by wanderlust.xx; 23-04-2012 at 14:33.
2. Re: Laplace transform, pretty quick (renewal process)
(Original post by wanderlust.xx)
I have a renewal function given as for .

The interarrival times are identically distributed with distribution function which is given by the equation .

I need to show that

Having solved the second order differential equation, I get that . Clearly I just need my constants A and B, but the only possible way I can do it straight from here is by using the fact that f is a density, so . I'm not sure what to do from here.
f isn't a density, it's a distribution function (at least that's what's written). Unfortunately this doesn't make sense. Have you quoted the question verbatim?
3. Re: Laplace transform, pretty quick (renewal process)
(Original post by Glutamic Acid)
f isn't a density, it's a distribution function (at least that's what's written). Unfortunately this doesn't make sense. Have you quoted the question verbatim?
Ah yes, sorry:

"Suppose that the common distribution function F of the interarrival times is repeatedly differentiable for t>0 and that f(t) = F'(t) is the common density function for t>0 of the interarrival times."

So it is indeed a density.
4. Re: Laplace transform, pretty quick (renewal process)
(Original post by wanderlust.xx)
Ah yes, sorry:

"Suppose that the common distribution function F of the interarrival times is repeatedly differentiable for t>0 and that f(t) = F'(t) is the common density function for t>0 of the interarrival times."

So it is indeed a density.
I don't think there's anything wrong with taking the Laplace transform of f and solving for A and B. You can substitute in a value of s which may make it easier (this, together with f being a density should give you what you need).
Last edited by Glutamic Acid; 23-04-2012 at 14:46.
5. Re: Laplace transform, pretty quick (renewal process)
(Original post by Glutamic Acid)
I don't think there's anything wrong with taking the Laplace transform of f and solving for A and B. You can substitute in a value of s which may make it easier (this, together with f being a density should give you what you need).
Hmm, I'm still hopelessly stuck.

Using the fact that f is a density, I've arrived at .

So

The laplace transform of the f I'm trying to find is

Substituting a value of 0 gives that .

Using the value of f I obtained from the differential equation, my transform was

But I don't see how we can use this to solve (i).

Oh wait, subbing 0 in should yield 1 since (i) is definitely correct, shouldn't it? Eurrghhh.
Last edited by wanderlust.xx; 23-04-2012 at 15:09.

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Last updated: April 23, 2012
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